Let $$ f_{\alpha}(x)=\phi_1(\alpha) \mathrm{e}^{-\frac{|x|^\alpha}{\phi_2(\alpha) }}, x \in \mathbb{R}, 0<\alpha<2, $$ where

  • $\phi_1(\alpha)=\frac{\alpha}{2} \left\{{\{\Gamma(3/\alpha)\}^{1/2}\over\{ \Gamma(1/\alpha)\}^{3/2}}\right\}$ and
  • $\phi_2(\alpha)= \left\{{\Gamma(1/\alpha)\over \Gamma(3/\alpha)}\right\}^{\alpha/2}$.

If $\it{F}_\alpha(\cdot)$ is a Fourier transform of $f_{\alpha}(x)$, show that $\it{F}_{\alpha_1}(\omega) \ge \it{F}_{\alpha_2}(\omega) $, for any $0<\alpha_1\le \alpha_2\le 2$ and $\omega\in \mathbb{R}$.

UPDATE. An equivalent problem is to show that Fourier transform of $$ (\exp(-|x|^\alpha \phi_3(\alpha)))(\phi_1’(\alpha) - |x|^{\alpha}\ln|x|\phi_1(\alpha)\phi_3(\alpha) - |x|^{\alpha}\phi_1(\alpha)\phi’_3(\alpha)), $$ $$\phi_3(\alpha)= \left\{{\Gamma(3/\alpha)\over \Gamma(1/\alpha)}\right\}^{\alpha/2}, $$ is nonpositive.
UPDATE TO UPDATE. Not quite an equivalent but may be somewhat easier problem. Let $$ f(x,\alpha)=\exp(-|x|B(\alpha))(|x|\ln |x|-A(\alpha)|x|+C(\alpha)) $$ where

  • $A(\alpha) = - \frac{\alpha}{2} \ln \left(\frac{\Gamma(3/\alpha)}{\Gamma(1/\alpha)}\right) + \frac{3}{2} \psi_0\left(\frac{3}{\alpha}\right)- \frac{1}{2} \psi_0\left(\frac{1}{\alpha}\right)$
  • $B(\alpha)=\left\{{\Gamma(3/\alpha)\over \Gamma(1/\alpha)}\right\}^{\alpha/2}$
  • $C(\alpha)= -\frac{1}{\phi_3(\alpha)} \left[1 - \frac{3}{2\alpha}\psi_0\left(\frac{3}{\alpha}\right) +\frac{3}{2\alpha}\psi_0\left(\frac{1}{\alpha}\right)\right]$ where $\psi_0(\cdot)$ is a digamma function.

Prove that Fourier transform of $f(x,\alpha)$ is positive for $0<\alpha<1$ or of $f(|x|^2,\alpha)$ for $0<\alpha<2$

  • $\begingroup$ does taking derivative $d(F_{\alpha}(\omega))/da$ lead to anything? Also I think in phi2 you meant to write $\alpha$ not p. $\endgroup$ Commented Apr 9, 2023 at 20:12
  • $\begingroup$ @ThomasKojar not really, the function gets very complicated. The original Fourier transform does not result in a closed form and so its derivative $\endgroup$ Commented Apr 9, 2023 at 20:15
  • $\begingroup$ I was more about thinking swapping derivative and integral. $\endgroup$ Commented Apr 9, 2023 at 20:16
  • $\begingroup$ Do you have any more background context on the the little $f_{\alpha}$ eg. literature/articles ? For example, do they satisfy any pde relation in $\alpha,x$? $\endgroup$ Commented Apr 9, 2023 at 20:19
  • $\begingroup$ @ThomasKojar yes I understood, the Fourier Transform of the derivative does not exist in a closed form . I did try to prove that the derivative has a nonnegative Fourier transform and failed ..Numerically yes the derivative has a positive Fourier transform. $\endgroup$ Commented Apr 9, 2023 at 20:20

1 Answer 1


Let me answer the "update to update" question:
Is the Fourier transform of $f(x,\alpha)=\exp(-|x|B(\alpha))(|x|\ln |x|-A(\alpha)|x|+C(\alpha))$ positive for $0<\alpha<2$?
The answer is no.

I compute the Fourier transform $F(t,\alpha)=\int_{-\infty}^\infty e^{ixt}f(x,\alpha)\,dx$ of $f(x,\alpha)$ at $t=0$, $$F(0,\alpha)=2\int_{0}^\infty e^{-B(\alpha)x} \bigl(x \ln x-A(\alpha) x+C(\alpha)\bigr)\,dx$$ $$\qquad=-2B(\alpha)^{-2} (A(\alpha)-B(\alpha) C(\alpha)+\ln B(\alpha)+\gamma_{\rm Euler} -1).$$ This is positive for $0<\alpha<1$ and negative for $1<\alpha<2$, see the plot.

For the record, if one wishes to check my algebra, I find $A(2)=3-\gamma_{\rm Euler} -\ln 2$, $B(2)=1/2$, $C(2)=1$, $F(0,2)=16 \ln 2-12$.
  • 1
    $\begingroup$ @TanyaVladi --- I am confused; you changed the question, didn't you? until an hour ago your question was "Prove that Fourier transform of $f(x,\alpha)$ is positive for $0<\alpha<2$" --- did I make a mistake there? $\endgroup$ Commented Apr 15, 2023 at 6:51
  • $\begingroup$ I made a mistake in my UPDATE UPDATE, stating something that is not equivalent. After you posted your answer hinting that my question is wrong people lost interest in it. Yes, I agree my UPDATE UPDATE is not equivalent to the original question. I based it on the fact that if $f(|x|)$ is positive definite so is $f(|x|^{\alpha})$, $0<\alpha<1$ but I am not sure about it either $\endgroup$ Commented Apr 15, 2023 at 15:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.